Question: Find $\lim_{x\to0^+}\left( x^{^{{{x}}^{{4}}}}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $64$ (Choice D) D The limit doesn't exist.
Answer: Substituting $x=0$ into $x^{{{x}}^{{4}}}$ results in the indeterminate form $0^{^{0}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=x^{{{x}}^{{4}}}$, we will find $\lim_{x\to 0^+}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to 0^+}y$. $\ln(y) =\dfrac{\ln(x)}{x^{-4}}$ Substituting $x=0$ into $\dfrac{\ln(x)}{x^{-4}}$ results in the indeterminate form $\dfrac{-\infty}{\infty}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to 0^+}\ln(y) \\\\ &=\lim_{x\to 0^+}\dfrac{\ln(x)}{x^{-4}} \\\\ &=\lim_{x\to 0^+}\dfrac{\dfrac{d}{dx}[\ln(x)]}{\dfrac{d}{dx}[x^{-4}]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0^+}\dfrac{\left(\dfrac{1}{x}\right)}{-4x^{-5}} \\\\ &=\lim_{x\to 0^+}\left(-\dfrac{x^{4}}{4}\right) \\\\ &=-\dfrac{0}{4} \gray{\text{Substitution}} \\\\ &=0 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0^+}\dfrac{\dfrac{d}{dx}[\ln(x)]}{\dfrac{d}{dx}[x^{-4}]}$ actually exists. We found that $\lim_{x\to 0^+}\ln(y)=0$, which means $\lim_{x\to 0^+}y=1$. [Why?] In conclusion, $\lim_{x\to0^+}\left( x^{^{{{x}}^{{4}}}}\right)=1$.